CH EM I STRY
11.
(a)
12.3 g of nitrobenzene is reduced to aniline
C6H5–NO2 + 6H + 6e–
C6H5 – NH2 + 3H2O
if the current efficiency for the process is 50% And the potential drop across the cell is 3V, how much
energy will be consumed in kJ. [moleclar weight of nitrobenzene = 123] [F = 96500 cb]
12.3
,
C6H5 – NH2 + 3H2O
50%
[
(b)
3V
= 123] [F = 96500 cb]
om
C 6H5–NO2 + 6H +
6e–
If an electron moving with a velocity 600 m s–1 which is accurate up to 0.005% then calculate uncertainty
in its position.
600 m s–1
w
w
w
.e
xa
m
ra
ce
.c
0.005%
[h = 6.63 × 10 –34 J s]
[mass of electron = 9.1 × 10–31 kg] (
Sol. (a)
C 6H5NO2 + 6H+ + 6e–
Moles of C6H5NO2 =
= 9.1 × 10–31 kg)
C6H5–NH2 + 3H2O
12.3
= 0.1 mol
123
1 mol C6H5NO2 requires 6F charge to redu e into niline
0.1 mol C6H5NO2 requires 0.6 F charge
red ce into aniline
Q = 0.6 F = 0.6 × 96500 cb
Current effeciency of proce s is 50%
required charge = 2 × (0.6 × 96500)c
= 115800 cb
Energy consume n the process = Q × V
= 115800 × 3 J
= 347400 J
= 347.4 KJ
(b)
–1
Given velocity o electron = 600 m s
uncertainty in velocity ( V) = 600 ×
0.005
= 30 × 10–3 m s–1
100
According to heisenberg's principle,
x. V
x
x
12.
(a)
4
h
4 m
3.14
x
6.63 10 34 J s
9.1 10 31kg 30
h
4 m. V
10 3ms
1
1.9 × 10 –3 m
Applying Bohr's model when H atom comes from n = 4 to n = 2, calculate its wavelength. In this process
write whether energy released or absorbed ? Also write the range of radiation ?
[RH = 2.18 × 10–28 J, h = 6.63 × 10–18 J]
H
n=4
[RH = 2.18 × 10–28 J, h = 6.63 × 10–18 J]
n=2
(b)
Write the electronic configuration of Gadolimum (Z = 64) and its most stable oxidation state.
(Z = 64)
Sol. *.[In this question value of h is incorrect]
According to Rydbeg's formula,
E=
hc
1 1
n12 n22
= RH Z 2
1 RH Z 2 1 1
hC n12 n22
where n1 = 2 and n2 = 4
2.18 10 18 J 1
1 1
-18
8
6.63 10 J 3 10 m / sec 4 16
6.63 10 18 J 3 10 8m / sec 16 318.24 10
3
6.54
2.18 10 18 J
(ii)
Energy is released
=
(iii)
= 48.66 × 108m/sec.
We can't discuss about range of radiation
stable oxidation state = +3
(b)
[Xe 54] 4f7 5d1 6s2
(a)
Explain the following (
)
w
w
w
.e
xa
m
ra
ce
.c
13.
8
om
(a)
(i)
+
+
Li ion move slowly than Cs ion in aqueous solution.
Li
Sol. (a)
+
(ii)
In crystal frenkel defect is not shown by Alkali mean halide bu silver halide shows.
(iii)
HF is less viscous than H2O.
HF, H2O
Draw all possible stereo isomers of complex [CO(en)2Cl2]
[CO(en)2Cl2]
(i)
+
+
Due to more hydrated ionic radii of Li it ecomes bulky in aqueous solution. i.e. while the mobility
+
+
of Li is less in comparison to Cs .
Li
+
Cs
+
(iii)
A molecule of H2O c n form four H-bonds while HF can form only two. (Or extent of H-bond in
H2O > HF)
H2C
C
NH2
CH2
Co
HC
C
CH2
NH2
H2C
+
H2N
H2N
H2C
+
NH2
C
H2C
C
CH2
Co
C
H2C NH2
Cis Leavo
Geometrical isomers
(i)
H2N CH2
Co
trans
(a)
+
In silver halide size of an on is much larger than cation.
H2N
14.
Li
(ii)
H2N
(b)
, Cs
+
(b)
+
C
NH2
NH2
CH2
H2N CH2
dextro
Enantiomers
(Mirror image)
optical isomers
Arrange the following in decreasing order of stability of dehydration.
OH
(I)
OH
(II)
OH
(III)
OH
(IV)
(iii)
CHO
CHO
CHO
CHO
CH3
NO2
Cl
OCH3
(I)
(II)
(III)
(IV)
Arrange the following in decreasing order of basic strength
H2O
OH
CH3 CO2
CH3 O
(I)
(II)
(III)
(IV)
(i)
Why amino acid is amphoteric in nature explain. (
(ii)
Name the polymer formed by adipic acid and hexamethylenediamine.
w
w
w
.e
xa
m
ra
ce
.c
(b)
Arrange the following in decreasing order of electrophilicity.
om
(ii)
Sol. (a)
(i)
I > IV > III > II
(ii)
II > III > I > IV
O
(b)
(i)
(iii)
IV > II > III
)
I
O
R–CH–C–OH
R–CH–C–O . In zwitter ionic form, amino acids show amphoteric
NH2
NH3
(Zwitter ion) behav our as they react both with acids and bases.
(ii)
15.
Nylon - 66
Explain the following (
(a)
(i)
)
Hydrolysis of CH3COCl is faster than CH3CONH2
CH3COCl
CH3 ONH
O
(i)
O
exists in Keto from but
O
(iii)
(b)
(i)
(i )
Sol. (a)
(i)
exists in enol form
O
( CH2 CH, CH C
CH2= CH is better nucleophile than CH C
Ph
H
C=C
H
HBr
CH3
O CO
NO2
Product, (
Product. (
)
)
Cl is better leaving group than NH2 .
O
(b)
)
while enol form of
O
is more stable due to aromatic nature.
(ii)
Keto form is more stable of
(iii)
CH2 –CH is stronger base than HC C and stronger base is better nucleophile.
(i)
Ph–CH–CH2CH 3
Br
(ii)
(a)
(b)
Sol.
(a)
O–C–
O
Cu metal crystallizes in face centred cubic lattice with cell edge, a = 361.6 pm. What is the density of Cu
23
crystal (Mwt. of Cu = 63.5 amu NA = 6.023 × 10 )
Complete the following reaction
–
(i)
O3 + I
(ii)
CaH2 + H2O
Number of atoms in Cu unit cell (n) = 4
Mwt. of Cu = 63.5 amu
3
–8
3
volume of unit cell (v) = a = (3.616 × 10 cm)
Crystal density =
n
Mwt.
v
NA
–
4
3.616
–8
10
63.5 g
cm
3
6.023
23
10
–
8.919 g / cm3
(b)
(i)
(ii)
H2O + O3 + I
CaH2 + 2H2O
(a)
(i)
What is the arrangement of atoms in the latice structure of di mond nd give contribution of each C
atom?
w
w
w
.e
xa
m
ra
ce
.c
I2 + 2OH + O2
Ca(OH)2 + 2H2
om
16.
NO2
17
(ii)
?
If Al3+ replaces Na+ at the edge centre of NaCl atice th n calculate vacancies in 1 mol NaCl.
Al3+
NaCl
Na
1 mol NaCl
?
(b)
(i)
2N 2O(g) + O2(g)
4NO(g) ;
H>0
What will be the effect on equilib ium when (
(ii)
(1)
Volume of the vessel increases (
(2)
Temperature decreases (
For the reaction (
2NO(g) + Cl2(g)
)
)
)
)
2NOCl(g)
If the pa tia pressure PNOCl = 0.32 atm, PNO = 0.22 atm and PCl = 0.11atm then find Kp ?
2
P NOCl = 0.32 atm, PNO = 0.22 atm
?
Sol. (a)
(i)
PCl = 0.11atm
2
Kp
The diamond lattice consist of a face centred cubic braviaus point latice which contains two identical
atoms per lattice point.
1
= 1 atom
8
1
= 3 atoms
At every face centre of unit cell = 6 ×
2
At alternative tetrahedral voids = 4 atoms, So total no. of atoms in an unit cell = 8 atoms
At every corner of unit cell = 8 ×
(ii)
Since in 1 unit cell of NaCl, 3 Na+ ions are present at the edge centres
So, 1 Al3+ will replace 3 Na+ ions.
i.e. 3/4 mol of Na+ ions will replaces by
1
mol of Al+3 (for electrically neutral)
4
and the remaining Na+ ions in crystal are
So in crystal
1
mol
4
1
1
1
1
mol of lattice sites will occupy by
mol ( mol Al+3 + mol Na+) of cations,
2
2
4
4
1
1
mol =
× 6.023 × 1023 = 3.0125 × 1023
2
2
therefore vacant sites (No. of cationic vacancies) =
(i)
(ii)
(1)
Equilibrium will be shifted in forward direction.
(2)
Equilibrium will be shifted in backward direction.
According to Law of mass action,
2
PNOCl
2
PNO.PCl2
Kp =
(0.32)2
= 19.23
(0.22) (0.11)
w
w
w
.e
xa
m
ra
ce
.c
O
om
(b)
18.
(a)
(i)
2 RLi
CH 3–C–OH
Cl
A
A
H3 O
B . Identify A and B (A
B
)
NO2
+ Na2CO3 + H2O
(ii)
C
Id ntify C. (C
)
NO2
(b)
Write the products formed by the reaction of glucose with (
Br 2 water (Br2
(i)
)
(ii) (Con
HNO3 (
OH
R
Sol. (a)
(i) CH3–C–OLi (A).
CH3–C–R (B)
OLi
O
19.
(a)
(ii)
(C)
(CH–OH) 4
CH2–OH
COOH
Gluconic a id
Glucaric acid
[Saccharic acid]
An LPG (Liqu fied petrolium gas) cylinder weighs 14.8 kg when empty. When full, it weights 29.0 kg and
shows a press e of 2.5 atm In the course of use at 27ºC, the weight of the full cylinder reduced to 23.2 kg.
Find out he volume of n-butane in cubic meters used up at 27ºC and 1 atm. [molecular wight of
n utane = 58]
LPG (
)
29.0 kg
14.8 kg
2.5 atm
27ºC
23.2 kg
n-
[n(b)
NO2
COOH
(CH–OH)4
(i)
HNO3)
NO2
COOH
(b)
(ii)
)
27ºC
1
= 58]
There is KI and sucrose solution with 0.1 M concentration, if the osmotic pressure of KI and sucrose
solution is 0.465 atm and 0.245 atm respectively. Then find the van't hoff factor of KI and its degree of
dissociation.
0.1 M
0.465
KI
KI
0.245
KI
?
Sol. (a)
Weight of n-butane used up = 29 – 23.2 = 5.8 kg
2.8 10 3
= 100 mol
58
Volume of n-butane used up
Moles of n-butane used up =
as
PV = nRT or V =
100 mol 0.0821 L - atm mol -1 300 K
= 2463 L = 2.463 m3
1 atm
V=
Osmotic pressure ( ) = iCRT ; i = Van't Hoff factor
For KI solution 0.465 = i × 0.1 × RT
om
(b)
nRT
P
...........(1)
For sucrose solution 0.245 = 1 × 0.1 × RT ..........(2) (i = 1)
w
w
w
.e
xa
m
ra
ce
.c
Equation (1) is divided by equation (2) gives
0.465
0.245
i,
i = 1.897
For KI, degree of dissociation is –
(i 1)
(n - 1)
=
20.
(a)
or
=
1.879 1
= 0.897
(2 – 1)
For first order raction if rate constant at 17ºC is 2 8 × 10
(b)
)
Calculate activation energy (R = 8 3 J/mol. K) (
0.10
–0.201ºC
(Kf = 1 86 K kg
(i)
( i)
k2
Ea
l g k =
2.
2
.
303
R
1
log
2.8 10 -4 s
1
2.8 10 -5 s
1
log 10 =
(b)
27ºC is 2.8 × 10–4 s–1
) (R = 8.3 J/mol. K)
Depression in freezing point o 0.10 molal solution of HF is –0.201ºC. Calculate percentage degree of
dissociation of HF (Kf = 1.86 K g mol–1)
HF
Sol. (a)
s– and at 27ºC is 2.8 × 10–4 s–1
Write the equation for calculation o ac ivation energy
(
(ii)
5
89.7%
2.8 × 10–5 s–1
17ºC
(i)
or
HF
mol–1)
1 1
T1 T2
Ea
2.303 8.3J/mol K
1
1
290K 300K
Ea
(300 290 )
J/mol or Ea= 166299.63 J/mol = 166.299 kJ/mol.
2.303 8.3 290 300
Tf = 0.201
observed Tf = i (molality × Kt) or 0.201 = i (0.1 × 1.86–1) or i = 1.0806
degree of dissociation of HF
=
i 1 (1.0806 )
= 0.0806 or
(n – 1)
(2 1)
= 8.06%